Let $(\mathbb{R}^n,L_n,\lambda _n)$ be a measure space and let $f:\mathbb{R}^n\rightarrow \mathbb{R}$ be continuous and bounded, then $\sup\lvert f(x)\rvert = \lVert f \lVert_\infty $ .
Proof: we have $\lvert f(x)\rvert \leq \lVert f \rVert_\infty $ everywhere because the set where $f(x)> \lVert f \rVert_\infty $ is open (by continuity of $f$) and negligible and so it is empty. Therefore $\sup\lvert f\rvert \leq \lVert f \rVert_\infty $ . Conversely, $\sup\lvert f\rvert$ is essential bound for $|f|$ so $\lVert f \rVert_\infty \leq \sup\lvert f\vert$ .$\square$
Only one thing I did't understand, which is why the set where $f(x)> \lVert f \rVert_\infty $ is negligible ??
($\lVert f \rVert_\infty = \inf\{ c/\lvert f(x)\rvert \leq c\; \text{almost everywhere}\}$)
It follows from your definition of $\|f\|_\infty$ that$$\|f\|_\infty=\inf\{x\in\Bbb R\mid f^{-1}([x,\infty))\text{ is negligible}\}.$$Therefore the set $\{x\in\Bbb R\mid f(x)>\| f\|_\infty\}$ is negligible, since it is equal to$$\bigcup_{n\in\Bbb N}f^{-1}\left(\left[\|f\|_\infty+\frac1n,\infty\right)\right)$$and each $f^{-1}\left(\left[\|f\|_\infty+\frac1n,\infty\right)\right)$ is negligible.